If $$ \sin A = \frac{4}{5} $$, then $$ (3 - an A)(2 + \cos A) $$ is:

Question

If $$ \sin A = \frac{4}{5} $$, then $$ (3 - 	an A)(2 + \cos A) $$ is:

Accepted Answer

Given $$ \sin A = \frac{4}{5} $$, we know that $$ \sin A = \frac{	ext{opposite}}{	ext{hypotenuse}} $$. Using the Pythagorean theorem, the adjacent side is calculated as:  
$$
\sqrt{5^2 - 4^2} = 3.
$$  
Thus, $$ \cos A = \frac{	ext{adjacent}}{	ext{hypotenuse}} = \frac{3}{5} $$ and $$ 	an A = \frac{	ext{opposite}}{	ext{adjacent}} = \frac{4}{3} $$.

Now, substituting these values into the given expression:  
$$
(3 - 	an A)(2 + \cos A) = (3 - \frac{4}{3})(2 + \frac{3}{5}).
$$

Simplify each term:  
$$
3 - \frac{4}{3} = \frac{9}{3} - \frac{4}{3} = \frac{5}{3},  
$$  
and  
$$
2 + \frac{3}{5} = \frac{10}{5} + \frac{3}{5} = \frac{13}{5}.
$$

Thus, the expression becomes:  
$$
\left(\frac{5}{3}ight)\left(\frac{13}{5}ight) = \frac{65}{15} = \frac{13}{3}.
$$

Therefore, the correct answer is $$ \frac{13}{3} $$.

Answer

Answer

Answer